Question

Write the resultant moment of the wrench. M =rc/pxFc+rg/p*F, +rp*F, M.uz = (xi + yj+0k)200j+((x-3)i+yj)+(-300k)+((x-3)i+(y-5)j)< 400i M.ur, = 200xk+((x-3)300j - 300 yi) - 400(y-5)k Mur = 200xk + 300xj - 900j - 300 yi - 400 yk + 2000k Substitute 0.7428i +0.3714j-0.557lk for us, Muy = 200.xk + 300xj - 900j-300 yi - 400 yk + 2000k M.(0.7428i +0.3714j-0.557 lk) = 200xk+ 300xj -900j - 300 yi - 400 yk + 2000k 0.7428 Mi+0.3714Mj -0.5571 Mk = i(-300y)+ j(300x - 900) +k (200x - 400 y + 2000) Equate icoefficient. 0.7428 M = -300 y v = 0.7428 300 ...... (1) Equate icoefficient. 0.3714M = 300x - 900 x=(900+0.3714M) ...... (2) Equate kcoefficient. -0.5571M = 200x - 400y + 2000 0.7428 11900+0.3714M). Substitute 300 for v and 30000+0.3714M) for x -0.5571M = 200(900 +0.3714M) 200 - 400 - 300 M +2000 -0.5571M = 600+0.2476M +0.9904M+2000 -1.7951M = 2600 M =-1448.387 N.m

this is in reply to the answer I was given for chapter 4 problem 137. Thank you so much for the very complete answer that you sent down and I just have one remaining question why are they position vectors inverted. I mean if point P is the origin and then say you direct that position vector down to see where it touches the line of force of C then is the position vector inverted as if it's coming from C back to P. That is my question otherwise thank you very much. if I go from point P to C then it is r pc < -x -y 0 >. However the first cross product indicates that the position vector is the inverse of that as if it is going from point C to point P which would be r c/p < x y 0 >. so that is my question why is the position vector inverted?

Answer 1

Yes you are correct. To get resultant moment about P, the axes are shifted with origin at P, hence the position vector directions will be from P towards A, B, and C.

So correct position vectors will be, r_C/P = -xi - yj + 0k

r_B/P = (3-x)i - yj + 0k

and r_A/P = (3-x)i + (5-y)j + 0k

Please replace position vectors with these correct values in above solution and then calculate.